Solution.
V(CaCl2)=30.0mL=0.03L;
c(CaCl2)=0.50mol/L;
a)c=Vν⟹ν=c⋅V;
ν(CaCl2)=0.50mol/L⋅0.03L=0.015mol;
c) 0.015mol xmol
CaCl2(aq)+2AgNO3(aq)−−>Ca(NO3)2(aq)+2AgCl(s);
1 2
x=10.015mol⋅2=0.03mol;
ν(AgCl)=0.03mol;
M(AgCl)=143.32g/mol;
ν=Mm⟹m=ν⋅M;
m(AgCl)=0.03mol⋅143.32g/mol=4.30g;
Answer: a) ν(CaCl2)=0.015mol;
c)m(AgCl)=4.3g.
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